無法獲得查詢結果顯示 - php/mysqli

<form action="send_post.php" method="post"> 
    <p>A: <input type="text" name="A"> 
    <p>B: <input type="text" name="B">          
    <p><input type="submit">  
</form>  
<?php  
    $db = "localhost"; $usr = "blah1"; $pword = "blah2"; $dbname = "blah3"; 
    -----connects to server ok -------    
    $A = mysqli_real_escape_string($conn, $_POST['A']); 
    $B = mysqli_real_escape_string($conn, $_POST['B']); 
    $query="SELECT * FROM blah4 where A=$A and B=$B";  
    $sql=mysqli_query($conn, $query);   
    echo $sql;             
    $conn->clos();                 
?>

需要什麼來顯示數據從數據庫？無法獲得查詢結果顯示 - php/mysqli

來源

2016-04-30 freebie.1

您需要顯示所有代碼。這對我們無能爲力。 – Matt

你應該[準備]（http://php.net/manual/en/mysqli.prepare.php）SQL層。比'mysqli_real_escape_string'更安全！ –

您的查詢字符串的格式不正確

<form action="send_post.php" method="post"> 
    <p>A: <input type="text" name="A"> </p> 
    <p>B: <input type="text" name="B"> </p>          
    <p><input type="submit"> </p> 
</form>  
<?php  
    $db = "localhost"; $usr = "blah1"; $pword = "blah2"; $dbname = "blah3"; 
    -----connects to server ok -------    
    $A = mysqli_real_escape_string($conn, $_POST['A']); 
    $B = mysqli_real_escape_string($conn, $_POST['B']); 
    $query="SELECT * FROM blah4 where A='$A' and B='$B'";  
    $sql=mysqli_query($conn, $query);   
    echo $sql;             
    $conn->close();                 
?>

或

$query="SELECT * FROM blah4 where A='". $A ,"' and B='" .$B ."'";

您還沒有關閉</p>

和

$conn->close();

來源

2016-04-30 19:37:02 scaisEdge

謝謝你，我已經設法解決了...... newbs，sheesh ... –

要添加在你沒有真正得到從DB數據試試這個

$row->mysqli_result::fetch_object($query);

那麼你可以得到某些領域，如如果u有一個用戶名字段

username = $row->username;

那麼你可以顯示用戶名像

echo $username;

來源

2016-04-30 19:42:50

無法獲得查詢結果顯示 - php/mysqli

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